# fun f z -> z;; - : 'a -> 'b -> 'b =Finally, we're getting consistent principle types, so we can stop. These types should remind you of the simply-typed lambda calculus types for Church numerals (`(o -> o) -> o -> o`) with one extra bit thrown in (in this case, an int). So here's our type constructor for our hand-rolled lists: type 'a list' = (int -> 'a -> 'a) -> 'a -> 'a Generalizing to lists that contain any kind of element (not just ints), we have type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b So an `('a, 'b) list'` is a list containing elements of type `'a`, where `'b` is the type of some part of the plumbing. This is more general than an ordinary OCaml list, but we'll see how to map them into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s in order to proceed to build a monad: l'_unit (x : 'a) : ('a, 'b) list = fun x -> fun f z -> f x z No problem. Arriving at bind is a little more complicated, but exactly the same principles apply, you just have to be careful and systematic about it. l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ... Unfortunately, we'll need to spell out the types: l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd) : ('c -> 'd -> 'd) -> 'd -> 'd = ... It's a rookie mistake to quail before complicated types. You should be no more intimiated by complex types than by a linguistic tree with deeply embedded branches: complex structure created by repeated application of simple rules. As usual, we need to unpack the `u` box. Examine the type of `u`. This time, `u` will only deliver up its contents if we give `u` as an argument a function expecting an `'a`. Once that argument is applied to an object of type `'a`, we'll have what we need. Thus: .... u (fun (a : 'a) -> ... (f a) ... ) ... In order for `u` to have the kind of argument it needs, we have to adjust `(f a)` (which has type `('c -> 'd -> 'd) -> 'd -> 'd`) in order to deliver something of type `'b -> 'b`. The easiest way is to alias `'d` to `'b`, and provide `(f a)` with an argument of type `'c -> 'b -> 'b`. Thus: l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b) : ('c -> 'b -> 'b) -> 'b -> 'b = .... u (fun (a : 'a) -> f a k) ... [Exercise: can you arrive at a fully general bind for this type constructor, one that does not collapse `'d`'s with `'b`'s?] As usual, we have to abstract over `k`, but this time, no further adjustments are needed: l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b) : ('c -> 'b -> 'b) -> 'b -> 'b = fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) -> f a k) You should carefully check to make sure that this term is consistent with the typing. Our theory is that this monad should be capable of exactly replicating the behavior of the standard List monad. Let's test: l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3] l'_bind (fun f z -> f 1 (f 2 z)) (fun i -> fun f z -> f i (f (i+1) z)) ~~># fun f z -> f 1 z;; - : (int -> 'a -> 'b) -> 'a -> 'b = # fun f z -> f 2 (f 1 z);; - : (int -> 'a -> 'a) -> 'a -> 'a = # fun f z -> f 3 (f 2 (f 1 z)) - : (int -> 'a -> 'a) -> 'a -> 'a =

let t1 = Node ((Node ((Leaf 2), (Leaf 3))), (Node ((Leaf 5),(Node ((Leaf 7), (Leaf 11)))))) . ___|___ | | . . _|__ _|__ | | | | 2 3 5 . _|__ | | 7 11Our first task will be to replace each leaf with its double:

let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) = match t with Leaf x -> Leaf (newleaf x) | Node (l, r) -> Node ((treemap newleaf l), (treemap newleaf r));;`treemap` takes a function that transforms old leaves into new leaves, and maps that function over all the leaves in the tree, leaving the structure of the tree unchanged. For instance:

let double i = i + i;; treemap double t1;; - : int tree = Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22))) . ___|____ | | . . _|__ __|__ | | | | 4 6 10 . __|___ | | 14 22We could have built the doubling operation right into the `treemap` code. However, because what to do to each leaf is a parameter, we can decide to do something else to the leaves without needing to rewrite `treemap`. For instance, we can easily square each leaf instead by supplying the appropriate `int -> int` operation in place of `double`:

let square x = x * x;; treemap square t1;; - : int tree =ppp Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))Note that what `treemap` does is take some global, contextual information---what to do to each leaf---and supplies that information to each subpart of the computation. In other words, `treemap` has the behavior of a reader monad. Let's make that explicit. In general, we're on a journey of making our treemap function more and more flexible. So the next step---combining the tree transducer with a reader monad---is to have the treemap function return a (monadized) tree that is ready to accept any `int->int` function and produce the updated tree. \tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))

\f . ____|____ | | . . __|__ __|__ | | | | f2 f3 f5 . __|___ | | f7 f11That is, we want to transform the ordinary tree `t1` (of type `int tree`) into a reader object of type `(int->int)-> int tree`: something that, when you apply it to an `int->int` function returns an `int tree` in which each leaf `x` has been replaced with `(f x)`. With previous readers, we always knew which kind of environment to expect: either an assignment function (the original calculator simulation), a world (the intensionality monad), an integer (the Jacobson-inspired link monad), etc. In this situation, it will be enough for now to expect that our reader will expect a function of type `int->int`.

type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *) let reader_unit (x:'a): 'a reader = fun _ -> x;; let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;It's easy to figure out how to turn an `int` into an `int reader`:

let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;; int2int_reader 2 (fun i -> i + i);; - : int = 4But what do we do when the integers are scattered over the leaves of a tree? A binary tree is not the kind of thing that we can apply a function of type `int->int` to.

let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader = match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x')) | Node (l, r) -> reader_bind (treemonadizer f l) (fun x -> reader_bind (treemonadizer f r) (fun y -> reader_unit (Node (x, y))));;This function says: give me a function `f` that knows how to turn something of type `'a` into an `'b reader`, and I'll show you how to turn an `'a tree` into an `'a tree reader`. In more fanciful terms, the `treemonadizer` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the monad through the leaves.

# treemonadizer int2int_reader t1 (fun i -> i + i);; - : int tree = Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))Here, our environment is the doubling function (`fun i -> i + i`). If we apply the very same `int tree reader` (namely, `treemonadizer int2int_reader t1`) to a different `int->int` function---say, the squaring function, `fun i -> i * i`---we get an entirely different result:

# treemonadizer int2int_reader t1 (fun i -> i * i);; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))Now that we have a tree transducer that accepts a monad as a parameter, we can see what it would take to swap in a different monad. For instance, we can use a state monad to count the number of nodes in the tree.

type 'a state = int -> 'a * int;; let state_unit x i = (x, i+.5);; let state_bind u f i = let (a, i') = u i in f a (i'+.5);;Gratifyingly, we can use the `treemonadizer` function without any modification whatsoever, except for replacing the (parametric) type `reader` with `state`:

let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state = match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x')) | Node (l, r) -> state_bind (treemonadizer f l) (fun x -> state_bind (treemonadizer f r) (fun y -> state_unit (Node (x, y))));;Then we can count the number of nodes in the tree:

# treemonadizer state_unit t1 0;; - : int tree * int = (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13) . ___|___ | | . . _|__ _|__ | | | | 2 3 5 . _|__ | | 7 11Notice that we've counted each internal node twice---it's a good excerice to adjust the code to count each node once. One more revealing example before getting down to business: replacing `state` everywhere in `treemonadizer` with `list` gives us

# treemonadizer (fun x -> [[x; square x]]) t1;; - : int list tree list = [Node (Node (Leaf [2; 4], Leaf [3; 9]), Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]Unlike the previous cases, instead of turning a tree into a function from some input to a result, this transformer replaces each `int` with a list of `int`'s. Now for the main point. What if we wanted to convert a tree to a list of leaves?

type ('a, 'r) continuation = ('a -> 'r) -> 'r;; let continuation_unit x c = c x;; let continuation_bind u f c = u (fun a -> f a c);; let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation = match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x')) | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x -> continuation_bind (treemonadizer f r) (fun y -> continuation_unit (Node (x, y))));;We use the continuation monad described above, and insert the `continuation` type in the appropriate place in the `treemonadizer` code. We then compute:

# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);; - : int list = [2; 3; 5; 7; 11]We have found a way of collapsing a tree into a list of its leaves. The continuation monad is amazingly flexible; we can use it to simulate some of the computations performed above. To see how, first note that an interestingly uninteresting thing happens if we use the continuation unit as our first argument to `treemonadizer`, and then apply the result to the identity function:

# treemonadizer continuation_unit t1 (fun x -> x);; - : int tree = Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))That is, nothing happens. But we can begin to substitute more interesting functions for the first argument of `treemonadizer`:

(* Simulating the tree reader: distributing a operation over the leaves *) # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) (* Simulating the int list tree list *) # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);; - : int list tree = Node (Node (Leaf [2; 4], Leaf [3; 9]), Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121]))) (* Counting leaves *) # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);; - : int = 5We could simulate the tree state example too, but it would require generalizing the type of the continuation monad to type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;